\(A=\dfrac{2}{\sqrt{17}+\sqrt{15}}\) ; \(B=\dfrac{2}{\sqrt{15}+\sqrt{13}}\)

Mà \(\sqrt{17}+\sqrt{15}>\sqrt{15}+\sqrt{13}>0\)

\(\Rightarrow\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{15}+\sqrt{13}}\)

\(\Rightarrow A< B\)

\(A=\sqrt{17}-\sqrt{15}=\dfrac{2}{\sqrt{17}+\sqrt{15}}\)

\(B=\sqrt{15}-\sqrt{13}=\dfrac{2}{\sqrt{13}+\sqrt{15}}\)

mà \(\dfrac{2}{\sqrt{17}+\sqrt{15}}< \dfrac{2}{\sqrt{13}+\sqrt{15}}\)

nên A<B

Lời giải:

a)

$(\sqrt{2}+\sqrt{3})^2=5+2\sqrt{6}=10-(2+3-2\sqrt{6})$

$=10-(\sqrt{2}-\sqrt{3})^2\leq 10$

$\Rightarrow \sqrt{2}+\sqrt{3}< \sqrt{10}$

b)

$\sqrt{15}.\sqrt{17}=\sqrt{15.17}=\sqrt{(16-1)(16+1)}=\sqrt{16^2-1}< \sqrt{16^2}=16$

c)

$(\sqrt{3}+2)^2=7+4\sqrt{3}$

$(\sqrt{2}+\sqrt{6})^2=8+4\sqrt{3}$

$\Rightarrow (\sqrt{3}+2)^2< (\sqrt{2}+\sqrt{6})^2$

$\Rightarrow \sqrt{3}+2< \sqrt{2}+\sqrt{6}$

d)

$(\sqrt{15}+\sqrt{17})^2=32+2\sqrt{15.17}< 32+2.16=64$ (theo kết quả câu b)

$\Rightarrow \sqrt{15}+\sqrt{17}< \sqrt{64}=8$

a) <

b) <

c) >

d) <

      a <

            b <

                           c >

                   d <

a/ \(\left(\sqrt{2}+\sqrt{3}\right)^2=2+3+2\sqrt{2.3}=5+2\sqrt{6}=5+\sqrt{24}\)

\(\left(\sqrt{10}\right)^2=10=5+5=5+\sqrt{25}\)

Vì \(\sqrt{24}< \sqrt{25}\)

=>\(\sqrt{2}+\sqrt{3}< \sqrt{10}\)

b/\(\left(\sqrt{3}+2\right)^2=3+4+4\sqrt{3}=7+4\sqrt{3}\)

\(\left(\sqrt{2}+\sqrt{16}\right)^2=2+16+2\sqrt{2.16}=18+4\sqrt{8}\)

=> \(\sqrt{3}+2< \sqrt{2}+\sqrt{16}\)

c/ \(16=\sqrt{16^2}\)

\(\sqrt{15}.\sqrt{17}=\sqrt{15.17}=\sqrt{\left(16-1\right)\left(16+1\right)}=\sqrt{16^2-1}\)

=> \(16>\sqrt{15}.\sqrt{17}\)

d/\(8^2=64=32+32=32+2\sqrt{256}\)

\(\left(\sqrt{15}+\sqrt{17}\right)^2=15+17+2\sqrt{15.17}=32+2\sqrt{255}\)

=> \(8>\sqrt{15}+\sqrt{17}\)

b: Ta có: \(4\sqrt{5}=\sqrt{4^2\cdot5}=\sqrt{80}\)

\(5\sqrt{3}=\sqrt{5^2\cdot3}=\sqrt{75}\)

mà 80>75

nên \(4\sqrt{5}>5\sqrt{3}\)

a)\(\sqrt{4}+\sqrt{14}=5,741657387\)

\(\sqrt{18}\)=4,242640687

->vay: dien dau >

b)\(\sqrt{15}+\sqrt{16}+\sqrt{17}+\sqrt{18}=16,23872966\)

\(\sqrt{90}=9,486832981\)

->vay : điền dấu <

a)\(\sqrt{4}+\sqrt{14}\) và \(\sqrt{18}\)

ta có : \(\sqrt{18}=\sqrt{14}+\sqrt{4}\)

suy ra : \(\sqrt{4}+\sqrt{14}=\sqrt{18}\)

b)\(\sqrt{15}+\sqrt{16}+\sqrt{17}+\sqrt{12}\)với \(\sqrt{90}\)

ta có :\(\sqrt{90}=\sqrt{20}+\sqrt{20}+\sqrt{20}+\sqrt{30}\)

mà :\(\sqrt{20}>\sqrt{15};\sqrt{20}>\sqrt{16};\sqrt{20}>\sqrt{17};\sqrt{30}>\sqrt{12}\)

suy ra :\(\sqrt{90}\)lớn hơn

Bài 1: 

Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)

Số giá trị nguyên thỏa mãn điều kiện là:

\(\left(2+4\right)+1=7\)